我希望有人可以帮助我解决我在特定项目中使用Eclipse中的java搜索功能的问题.
在一个特定项目上使用java搜索时,我收到一条错误消息Class file name must end with .class
(请参阅下面的堆栈跟踪).这似乎并没有发生在所有项目上,只是一个特定项目,所以也许我应该尝试重建一些东西?
我已经尝试过Project -> Clean
...并关闭Eclipse,删除所有构建的类文件并重启Eclipse无济于事.
我在谷歌上找到的唯一一个问题是http://www.crazysquirrel.com/computing/java/eclipse/error-during-java-search.jspx,但不幸的是他的解决方案(关闭,删除类文件,重新启动)对我不起作用.
如果有人可以提出尝试的建议,或者我可以收集任何可能有助于跟踪它的信息,我会非常感谢指针.
Version: 3.4.0
Build id: I20080617-2000
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还发现了这个帖子 - http://www.myeclipseide.com/PNphpBB2-viewtopic-t-20067.html - 这表明当项目名称包含句点时可能会出现同样的问题.不幸的是,在我的设置中并非如此,所以我仍然卡住了.
Caused by: java.lang.IllegalArgumentException: Class file name must end with .class
at org.eclipse.jdt.internal.core.PackageFragment.getClassFile(PackageFragment.java:182)
at org.eclipse.jdt.internal.core.util.HandleFactory.createOpenable(HandleFactory.java:109)
at org.eclipse.jdt.internal.core.search.matching.MatchLocator.locateMatches(MatchLocator.java:1177)
at org.eclipse.jdt.internal.core.search.JavaSearchParticipant.locateMatches(JavaSearchParticipant.java:94)
at org.eclipse.jdt.internal.core.search.BasicSearchEngine.findMatches(BasicSearchEngine.java:223)
at org.eclipse.jdt.internal.core.search.BasicSearchEngine.search(BasicSearchEngine.java:506)
at org.eclipse.jdt.core.search.SearchEngine.search(SearchEngine.java:551)
at org.eclipse.jdt.internal.corext.refactoring.RefactoringSearchEngine.internalSearch(RefactoringSearchEngine.java:142)
at org.eclipse.jdt.internal.corext.refactoring.RefactoringSearchEngine.search(RefactoringSearchEngine.java:129)
at org.eclipse.jdt.internal.corext.refactoring.rename.RenameTypeProcessor.initializeReferences(RenameTypeProcessor.java:594)
at org.eclipse.jdt.internal.corext.refactoring.rename.RenameTypeProcessor.doCheckFinalConditions(RenameTypeProcessor.java:522)
at org.eclipse.jdt.internal.corext.refactoring.rename.JavaRenameProcessor.checkFinalConditions(JavaRenameProcessor.java:45)
at org.eclipse.ltk.core.refactoring.participants.ProcessorBasedRefactoring.checkFinalConditions(ProcessorBasedRefactoring.java:225)
at org.eclipse.ltk.core.refactoring.Refactoring.checkAllConditions(Refactoring.java:160)
at org.eclipse.jdt.internal.ui.refactoring.RefactoringExecutionHelper$Operation.run(RefactoringExecutionHelper.java:77)
at org.eclipse.jdt.internal.core.BatchOperation.executeOperation(BatchOperation.java:39)
at org.eclipse.jdt.internal.core.JavaModelOperation.run(JavaModelOperation.java:709)
at org.eclipse.core.internal.resources.Workspace.run(Workspace.java:1800)
at org.eclipse.jdt.core.JavaCore.run(JavaCore.java:4650)
at org.eclipse.jdt.internal.ui.actions.WorkbenchRunnableAdapter.run(WorkbenchRunnableAdapter.java:92)
at …
Run Code Online (Sandbox Code Playgroud) 我有一个大的xlsx文件,我必须实现搜索选项的大量数据我使用了Apache POI jar以及jxl jar,以便在行和列之间进行搜索.但是需要花费大量时间来遍历大数据,有些人可以帮助我,这是任何jar文件或任何其他可用于在Excel文件上更快地进行搜索的概念......
String searchValue="my_value_to_search";
for (int i = 0; i < sheet.getColumns(); i++) {
for (int j = 0; j < sheet.getRows(); j++) {
value = sheet.getCell(i, j);
valueType = value.getType();
String val=getCellType(valueType, value);
if (val != null&&val==searchValue) {
// To do manipulation.
}
}
}
Run Code Online (Sandbox Code Playgroud) 我面临以下错误:
我编写了一个基于 Spark 流 ( Dstream )的应用程序来提取来自 PubSub 的消息。不幸的是,我在执行这项工作时遇到了错误。实际上,我正在使用由 4 个节点组成的集群来执行 spark 作业。
在没有任何特定错误的情况下运行作业 10 分钟后,我永久收到以下错误:
错误 org.apache.spark.streaming.CheckpointWriter:
无法将检查点任务提交给线程池执行程序 java.util.concurrent.RejectedExecutionException:任务 org.apache.spark.streaming.CheckpointWriter$CheckpointWriteHandler@68395dc9
从 java.util.concurrent被拒绝.ThreadPoolExecutor@1a1acc25
[正在运行,池大小 = 1,活动线程 = 1,排队任务 = 1000,已完成任务 = 412]
我们需要appednd ORC
文件.我试着谷歌但没有结果.另外org.apache.hadoop.hive.ql.io.orc.WriterImpl
的ORC
没有追加API.无论如何都要附加ORC文件?(更具体地说,使用JAVA)
我在一个消费者的hornetQ中创建了会话,然后我使用producer在队列中添加了4个消息.在此之后,我创造了新的消费者.
这位消费者会不会了解旧消息?
如果不是,是否可以在XML中配置它?
我创建了一个无法获取以前消息的新消费者.我只想确认这种行为是否正确?我没有在文档中找到任何帮助.
以下是代码段:
TextMessage receivedMessage = (TextMessage)consumer.receive();
receivedMessage.acknowledge();
System.out.println("Got order: " + receivedMessage.getText());
//consumer.close();
MessageConsumer newConsumer = session.createConsumer(orderQueue);
receivedMessage = (TextMessage)newConsumer.receive();
receivedMessage.acknowledge();
System.out.println("Got order: " + receivedMessage.getText());
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如果我取消注释consumer.close()行,它可以正常工作.
我的hornetq-jms.xml
<connection-factory name="NettyConnectionFactory">
<xa>true</xa>
<connectors>
<connector-ref connector-name="netty"/>
</connectors>
<entries>
<entry name="/XAConnectionFactory"/>
</entries>
<consumer-window-size>0</consumer-window-size>
</connection-factory>
<connection-factory name="NettyConnectionFactory">
<xa>false</xa>
<connectors>
<connector-ref connector-name="netty"/>
</connectors>
<entries>
<entry name="/ConnectionFactory"/>
</entries>
<consumer-window-size>0</consumer-window-size>
</connection-factory>
<connection-factory name="NettyThroughputConnectionFactory">
<xa>true</xa>
<connectors>
<connector-ref connector-name="netty-throughput"/>
</connectors>
<entries>
<entry name="/XAThroughputConnectionFactory"/>
</entries>
<consumer-window-size>0</consumer-window-size>
</connection-factory>
<connection-factory name="NettyThroughputConnectionFactory">
<xa>false</xa>
<connectors>
<connector-ref connector-name="netty-throughput"/>
</connectors>
<entries>
<entry name="/ThroughputConnectionFactory"/> …
Run Code Online (Sandbox Code Playgroud) 我有如下所示的文件:
{
name: "testing",
place:"London",
documents: [
{
x:1,
y:2,
},
{
x:1,
y:3,
},
{
x:4,
y:3,
}
]
}
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我想检索所有匹配的文档,即我想要以下格式的 o/p:
{
name: "testing",
place:"London",
documents: [
{
x:1,
y:2,
},
{
x:1,
y:3,
}
]
}
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我尝试过的是:
db.test.find({"documents.x": 1},{_id: 0, documents: {$elemMatch: {x: 1}}});
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但是,它只提供第一个条目。
由于缺乏信息,我发布的早期问题已经结束.
如果我在这里遗漏了什么,请告诉我.变换器似乎是将文件:/添加到文件路径的开头.
我在Solaris环境中工作,以下是应用转换时会发生的情况:
DOMSource sourcexml = new DOMSource(doc);
StreamResult resultxml = new StreamResult(new File("file.xml"));
transformer.transform(sourcexml, resultxml);
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我得到的例外是:
javax.xml.transform.TransformerException: java.io.FileNotFoundException: file:/opt/origenate/or_dev87/apps/documentarchive/file.xml (No such file or directory)
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注意,该文件存在于/opt/origenate/or_dev87/apps/documentarchive/file.xml中,但变换器对象正在查找文件:/opt/origenate/or_dev87/apps/documentarchive/file.xml.
为什么要附加文件:/?无论如何我可以删除它吗?
我是 MongoDB 的新手。我以以下格式将数据存储在 mongoDB 中
"_id" : ObjectId("51d5725c7be2c20819ac8a22"),
"chrom" : "chr22",
"pos" : 17060409,
"information" : [
{
"name" : "Category",
"value" : "3"
},
{
"name" : "INDEL",
"value" : "INDEL"
},
{
"name" : "DP",
"value" : "31"
},
{
"name" : "FORMAT",
"value" : "GT:PL:GQ"
},
{
"name" : "PV4",
"value" : "1,0.21,0.00096,1"
}
],
"sampleID" : "Job1373964150558382243283"
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我想将值更新为 11,其名称为Category
. 我试过下面的查询:
db.VariantEntries.update({$and:[ { "pos" : 117199533} , { "sampleID" : "Job1373964150558382243283"},{"information.name":"Category"}]},{$set:{'information.value':'11'}})
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但是 Mongo 回复了
can't …
Run Code Online (Sandbox Code Playgroud) 我有如下代码
//Create acm certificate for livy_cert
resource "aws_acm_certificate" "livy_cert" {
count = local.count
domain_name = "${var.subsystem}-${var.component}-livy.${var.region_fqdn}"
validation_method = "DNS"
lifecycle {
create_before_destroy = true
}
}
//Validation route53
resource "aws_route53_record" "certificate_validation" {
for_each = {
for dvo in aws_acm_certificate.livy_cert[0].domain_validation_options : dvo.domain_name => {
name = dvo.resource_record_name
record = dvo.resource_record_value
type = dvo.resource_record_type
}
}
name = each.value.name
records = [each.value.record]
ttl = 60
type = each.value.type
zone_id = module.core_info.route53_zone_id
}
//Validate certificate before assigning
resource "aws_acm_certificate_validation" "livy_alb_validation_cert" {
count = …
Run Code Online (Sandbox Code Playgroud) private class HandlerClass implements ActionListener
{
public void actionPerfomed(ActionEvent event)
{
JOptionPane.showMessageDialog(null, String.format("%s", event.getActionCommand));
}
}
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这是我的代码的一部分,当我编译时,我得到一个错误,说HandlerClass不是抽象的,不会覆盖抽象方法actionPerformed(java.awt.event.ActionEvent) in java.awt.event.ActionListener
.据我所知,actionPerformed
应该覆盖HandlerClass
不应该吗?我已经尝试abstract
在单词class之前添加" "但是我得到了另一个错误,因为我无法调用抽象类.我不确定是否可以用一个例外来解决这个问题?
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